# Permuations of identical items

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1. Nov 7, 2016

### mr.tea

1. The problem statement, all variables and given/known data
There is a book with 2 volumes. Each volume exists in 3 different languages. Each language has 2 identical copies(total of 12 books).
In how many ways we can arrange them on a shelf, with no restrictions and order of the volumes is irrelevant?

2. Relevant equations

3. The attempt at a solution
I thought that since we have 12 books, we can arrange them in $\frac{12!}{(2!)^6}$ but apparently it is not correct. The correct answer is $\frac{12!}{6!}$ but I am not sure why.

What am I missing?
Thank you.

Last edited: Nov 7, 2016
2. Nov 7, 2016

### PeroK

You're missing an attempt at the solution, so we can see where you go wrong.

3. Nov 7, 2016

### PeroK

PS as the question is stated, I would say the answer is 1. If order is irrelevant, then you have the same 12 books on the shelf in all cases.

4. Nov 7, 2016

### PeroK

PPS I agree with your answer! If you assume that order is relevant.

5. Nov 7, 2016

### mr.tea

I have just edited that the order of the volumes is irrelevant... does it add any information?

6. Nov 7, 2016

### PeroK

That doesn't make a lot of sense to me. If books are on a shelf, then the order defines the arrangement; otherwise, as I said, you only have the one set of books. It's different if two books are identical, then you can swap those without changing the arrangement. But, if you swap Volume 1 and Volume 2, by definition you have a different arrangement of books.

I'd be happy that you solved the question the sensible way. I wouldn't get too hung up on what precisely the question setter had in mind in this case.

7. Nov 7, 2016

### mr.tea

Well, we have 6 sets of books, each set consists of 2 identical books. Like: $A_1,A_1;B_1,B_1;C_1,C_1;A_2,A_2;B_2,B_2;C_2,C_2;$ where the 1's and 2's refer to the volume(vol1 or vol2). That's the case the question talks about. So I can't understand why my answer is not correct... Maybe it's not me?

8. Nov 7, 2016

### PeroK

This is why you should have posted your working in the first place. You can swap the two $A_1$'s etc. without changing the order. That's why I thought you got $\frac{12!}{(2!)^6}$.

I can't make sense of the order of volumes not being relevant. If you interpret that as $A_1, A_1, A_2, A_2$ can be in any order, then you'd just get $\frac{12!}{(4!)^3}$.